Finding lim x → ∞ ( 2 x − 5 ) 4 ( 2 x 2 + 1 ) ( 3 x 2 − 2 )

Question

Finding      lim          x      →      ∞                  (      2      x      −      5              )        4                    (      2              x        2            +      1      )      (      3              x        2            −      2      )

komizmtk · Accepted Answer

The dominant term is       x    4   (Imagine that you expanded both the numerator and denominator. What would the highest power of x be?). So, you could either multiply numerator and denominator by   1      /        x    4   (see below), or factor as follows:                                                        (              2              x              −              5                              )                4                                                    (              2                              x                2                            +              1              )              (              3                              x                2                            −              2              )                                                  =                                                                              (                                            x              (              2              −                                                5                  x                                            )                                                                    )                                                  4                                                                    x                2                            (              2              +                                                1                                      x                    2                                                              )              ⋅                              x                2                            (              3              −                                                2                                      x                    2                                                              )                                                                =                                                            x                4                            (              2              −                                                5                  x                                                            )                4                                                                    x                2                            (              2              +                                                1                                      x                    2                                                              )              ⋅                              x                2                            (              3              −                                                2                                      x                    2                                                              )                                                                =                                            (              2              −                                                5                  x                                                            )                4                                                    (              2              +                                                1                                      x                    2                                                              )              (              3              −                                                2                                      x                    2                                                              )                                      .            The limit as x tends to infinity is                    2        4                    2        ⋅        3              =  16      /    6  =  8      /    3.The   (  2  −  5      /    x      )    2   is probably a typo..Using the other approach, where you multiply numerator and denominator by   1      /        x    4  In the numerator, to distribute   1      /        x    4   over   (  2  x  −  5      )    4                      1                  x          4                      (    2    x    −    5          )      4        =                  (                            1        x              (    2    x    −    5    )                            )                    4        =    (    2    −                  5        x                    )      4        .  (just using       a    n        b    n    =  (  a  b      )    n   here).In the denominator, you could do the following:                    1                  x          4                      (    2          x      2        +    1    )    (    3          x      2        −    2    )    =                  1                  x          2                      (    2          x      2        +    1    )                  1                  x          2                      (    3          x      2        −    2    )    =    (    2    +                  1                  x          2                      )    (    3    −                  2                  x          2                      )  (or just expand   (  2      x    2    +  1  )  (  3      x    2    −  2  ) first and then distribute the   1      /        x    4   across).

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